\( \int_{0}^{1}{\rm erf^2}(x)\,dx \)

[Tolaso J Kos]

Evaluate : \( \displaystyle \int_{0}^{1}{\rm erf^2}(x)\,dx \).

[Papapetros Vaggelis]

Hello Tolis.

The function \(\displaystyle{\rm{erf}}\) is well defined, continuous and analytic on \(\displaystyle{\mathbb{C}}\) and given

by \(\displaystyle{\rm{erf}(z)=\dfrac{2}{\sqrt{\pi}}\,\int_{0}^{z}e^{-t^2}\,\mathrm{d}t}\) , cause the function \(\displaystyle{t
\mapsto e^{-t^2}\,,t\in\mathbb{C}}\)

is continuous and analytic on \(\displaystyle{\mathbb{C}}\) , with \(\displaystyle{\left(\rm{erf}(z)\right)'=\dfrac{2}{\sqrt{\pi}}\,e^{-z^2}\,,z\in\mathbb{C}}\) .

By applying integration by parts, we get :

\(\displaystyle{\begin{aligned} \int_{0}^{1}\rm{erf}^2\,(x)\,\mathrm{d}x&=\left[x\,\rm{erf}^2\,(x)\right]_{0}^{1}-\frac{4}{\pi}\int_{0}^{1}2\,x\,\rm{erf}\,(x)\,\left(\rm{erf}\,(x)\right)'\,\mathrm{d}x\\&=\rm{erf}^2\,(1)-\frac{4}{\pi}\int_{0}^{1}\rm{erf}\,(x)\,\mathrm{d}\,(e^{-x^2})\\&=\rm{erf}^2\,(1)-\frac{4}{\pi}\left[\rm{erf}\,(x)\,e^{-x^2}\right]_{0}^{1}+\frac{4}{\pi}\int_{0}^{1}e^{-2\,x^2}\,\mathrm{d}x\\&=\rm{erf}^2\,(1)-\frac{4}{\pi}e^{-1}\,\rm{erf}\,(1)+\sqrt{\frac{2}{\pi}}\int_{0}^{1}e^{-(\sqrt{2}\,x)}\,\mathrm{d}x\\&=\rm{erf}^2\,(1)-\frac{4}{\pi}e^{-1}\,\rm{erf}\,(1)+\sqrt{\frac{2}{\pi}}\rm{erf}\,(\sqrt{2})\end{aligned}}\)